Neural Sync Active

jan-2026-python-week-7-strings-1

Registry Synced

jan-2026-python-week-7-strings-1

504 words

3 min read

Strings 1

Course: Jan 2026 - Python

Strings 1

Submission deadline has passed for this assignment

Due Apr 01, 2026 at 11:59 PM IST

Instructions

Use "Test Run" to verify your code with public test cases.

Press "Submit" to have your assignment evaluated.

You can submit your assignment multiple times up until the deadline.

Make sure to submit your final code by the deadline to receive your score.

Summary

100 out of100

Score

Public Tests

2/2 Passed

Submitted on Apr 01, 2026 at 3:48 AM IST

Private Tests

2/2 Passed

Submitted on Apr 01, 2026 at 3:48 AM IST

**Change in eligibility criteria to write oppe2 exam: A5>=40/100 AND A6>=40/100 AND A7>=40/100 AND A8>=40/100. and becoming eligible to give the end term exam.

Check if all the odd indices are alphabets and even indices are digits**

Given a string, check if all the odd indices are alphabets and the even indices are digits.

Note: indices starts from 0.

Template Code(Click to Expand)



def is_odd_indices_alpha_and_even_indices_digits(string: str) -> bool:
    '''
    Given a string, check if all the odd indices are alphabets and the even indices are digits.

    Note: indices starts from 0.

    Arguments:
    string: str - the input string

    Return:
    bool - True if all odd indices are alphabets and even indices are digits, else False
    '''
    ...

NOTE: You can use the below tools for working out and debugging. Click to open them in new tab.

PythonTutor | Starboard Notebook | Pyodide Terminal

Public Tests ( 2/2 )

Case 1

Input:


text
is_equal(
    is_odd_indices_alpha_and_even_indices_digits("a1b2c3"),
    False
)
is_equal(
    is_odd_indices_alpha_and_even_indices_digits("1a2b3c"),
    True
)

Case 1

Expected Output:


text
False

True

Case 1

Actual Output:


text
False

True

Case 2

Input:


text
is_equal(
    is_odd_indices_alpha_and_even_indices_digits("3x4b6d"),
    True
)
is_equal(
    is_odd_indices_alpha_and_even_indices_digits("123abc"),
    False
)

Case 2

Expected Output:


text
True

False

Case 2

Actual Output:


text
True

False

Private Tests ( 2/2 )

Case 1

Input:


text
is_equal(is_odd_indices_alpha_and_even_indices_digits("0b1h2t"), True)
is_equal(is_odd_indices_alpha_and_even_indices_digits("4bn4n5"), False)

Case 2

Input:


text
is_equal(is_odd_indices_alpha_and_even_indices_digits("jqn432p5n6"), False)
is_equal(is_odd_indices_alpha_and_even_indices_digits("0a1b2d3e"), True)
is_equal(is_odd_indices_alpha_and_even_indices_digits("23l45k"), False)

Case 2

Expected Output:


text
False

True

False

Case 2

Actual Output:


text
False

True

False

💻 IITM Official Solution


python

def is_odd_indices_alpha_and_even_indices_digits(string: str) -> bool:
    '''
    Given a string, check if all the odd indices are alphabets and the even indices are digits.

    Note: indices starts from 0.

    Arguments:
    string: str - the input string

    Return:
    bool - True if all odd indices are alphabets and even indices are digits, else False
    '''

    return string[::2].isdigit() and string[1::2].isalpha()

💻 My Submitted Code


python
def is_odd_indices_alpha_and_even_indices_digits(string: str) -> bool:
    # Separate even-indexed and odd-indexed characters using slicing
    even_chars = string[0::2]
    odd_chars = string[1::2]

    # .isdigit() returns True if all characters in the string are digits
    # .isalpha() returns True if all characters in the string are alphabets
    # string[0::2] checks indices 0, 2, 4...
    # string[1::2] checks indices 1, 3, 5...
    return even_chars.isdigit() and odd_chars.isalpha()